How can we derive all of the continuous functions $f \colon \mathbb{R} \to \mathbb{R}$ such that $V=\{af_b : a,b \in \mathbb{R}\}$ is a vector space of dimension two, where $f_b\colon \mathbb{R} \to \mathbb{R}$ is defined as the translate $f_b(x)=f(x+b)$?

An example is $f=\sin$. The set $V$ contains the linearly independent functions $\sin$ and $\cos$ (since $\cos(x)=\sin(x+\frac{\pi}{2})$). By elementary trigonometry we have $c\sin(x+\phi) = s\sin x + t\cos x$ for any $c, \phi \in \mathbb{R}$ and some $s,t \in \mathbb{R}$, so it follows that $V$ is a vector space of dimension two spanned by $\sin$ and $\cos$.

This question has been answered in the case that $f$ is differentiable – $f$ is any solution of the ODE $f''+Bf'+Cf=0$ besides $ce^{rx}$ – but are there other non-differentiable continuous solutions to the problem, or are all the solutions differentiable?


All continuous solutions are differentiable.

Let $\mathcal{C}$ be the space of continuous functions over $\mathbb{R}$. Let $f \in \mathcal{C}$ be a function such that $V = \langle f_b : b \in \mathbb{R} \rangle$, the span of $f_b$ as a subspace of $\mathcal{C}$ is two dimenisonal. Let $$B = \bigg\{ b \in \mathbb{R} : f, f_b \text{ linear dependent on }\mathcal{C} \bigg\}$$ It is clear $B$ is a subgroup of $\mathbb{R}$ for addition. If $0 \in B$ is not an isolated point of $B$, then it is easy to see $B$ is dense in $\mathbb{R}$. Using continuity of $f_b$ w.r.t $b$, it is not hard to verify $f$ is an exponential function and leads to a contradiction that $V$ is one dimensional. As a result, $0$ is an isolated point of $B$.

WOLOG, we will assume $B \cap (-L,L) = \{ 0 \}$ for some $L > 1$ and take $f$ and $f_1$ as a basis of $V$. Let $A : \mathbb{R} \to M_{2}(\mathbb{R})$ be the $2 \times 2$ matrix valued function on $\mathbb{R}$ defined by:

$$\begin{align} \begin{pmatrix}f_{b}\\f_{b+1}\end{pmatrix} = A(b) \begin{pmatrix}f\\f_1\end{pmatrix} \end{align}\tag{*1}$$ Since $f$ and $f_1$ is linear independent, there are $x, y \in \mathbb{R}$ such that the matrix $\begin{pmatrix}f(x) & f(y)\\f_1(x)&f_1(y)\end{pmatrix}$ is invertible. From $(*1)$, we have $$\begin{pmatrix}f_b(x)&f_b(y)\\f_{b+1}(x)&f_{b+1}(y)\end{pmatrix} = A(b) \begin{pmatrix}f(x) & f(y)\\f_1(x)&f_1(y)\end{pmatrix}$$

Since the LHS is a continuous function in $b$, so does $A(b)$. For any $b, c \in \mathbb{R}$, it is clear $A$ satisfies:

$$\begin{cases} & A(b) A(c) = A(b+c) = A(c)A(b)\\ \text{ and } & A(b) A(-b) = A(-b)A(b) = A(0) = I_2\tag{*2} \end{cases}$$

This implies $A(b)$ is always invertible.

Since $A(b)$ is continuous in $b$ and $A(0) = I_2$, there exists a $\epsilon > 0$ such that $\| A(b)-I_2 \| < \frac12$ for $b \in (-\epsilon, \epsilon)$. This implies

$$\left\| \int_0^b \left( A(t) - I_2 \right) dt \right\| \le \int_0^b \left\| A(t) - I_2 \right\| dy < \frac{b}{2}\quad\text{ for }b \in (0,\epsilon)$$

As a result, $\quad\displaystyle\int_0^b A(t) dt = b I_2 + \int_0^b \left(A(t) - I_2\right) dt\quad$ is invertible over $(0,\epsilon)$.

Let $D : (0,\epsilon) \to M_2(\mathbb{R})$ be the function $\quad\displaystyle D(b) = \left( A(b) - 1 \right)\left( \int_0^b A(t) dt \right)^{-1}$.
For any $b,c \in (0,\epsilon)$, we have

$$\int_0^{b+c} A(t) dt = \left( \int_0^b + \int_b^{b+c} \right) A(t) dt = I_2 \left( \int_0^b A(t) dt \right) + A(b) \left( \int_0^c A(t) dt \right)$$ Exchange the role of $b, c$ and subtract, we get $$ \begin{align} & \left(A(b) - I_2\right)\left( \int_0^c A(t) dt \right) -\left(A(c) - I_2\right)\left( \int_0^b A(t) dt \right) = 0\\ \implies & \left(D(b) - D(c)\right) \left( \int_0^b A(t) dt \right) \left( \int_0^c A(t) dt \right) = 0\\ \implies & D(b) = D(c) \end{align}$$

i.e. $D(b)$ is a constant matrix over $(0,\epsilon)$. We will use the same $D$ to denote this constant matrix. Over $(0,\epsilon)$, $A(b)$ satisfies an integral equation

$$A(b) = I_2 + D \int_0^b A(t) dt$$ and hence $A(b) = e^{D b}$ there. Using $(*2)$, we see this expression is valid for all $b$. From this, we can conclude $A$ and hence $f$ is differentiable.

  • $\begingroup$ Brilliant! Thank you for answering. $\endgroup$ – Malper Jan 1 '14 at 6:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.